1 Introduction
A colouring of a graph is a labelling of the vertices so that adjacent vertices do not have the same label. We call these labels colours. Graph colouring problems are central to the study of combinatorial algorithms and they have many theoretical and practical applications. A typical problem asks whether a colouring exists under certain constraints, or how difficult it is to find such a colouring. For example, in the List Colouring problem, a graph is given where each vertex has a list of colours and one wants to know if the vertices can be coloured using only colours in their lists. The Choosability problem asks whether such list colourings are guaranteed to exist whenever all the lists have a certain size. In fact, an enormous variety of colouring problems can be defined and there is now a vast collection of literature on this subject. For longer introductions to the type of problems we consider we refer to two recent surveys [9, 16].
In this paper, we are concerned with the computational complexity of colouring problems. For many such problems, the complexity is well understood in the case where we allow every graph as input, so it is natural to consider problems with restricted inputs. We consider a variant of the List Colouring problem, closely related to Choosability, and give a complete classification of its complexity for planar graphs and a number of subclasses of planar graphs by combining known results with new results. Some of the known results are for (planar) graphs with bounded degree. We use these results to fill some more complexity gaps by giving a complete complexity classification of a number of colouring problems for graphs with bounded maximum degree.
1.1 Terminology
A colouring of a graph is a function such that whenever . We say that is the colour of . For a positive integer , if for all then is a colouring of . We say that is colourable if a colouring of exists. The Colouring problem is to decide whether a graph is colourable for some given integer . If is fixed, that is, not part of the input, we obtain the Colouring problem.
A list assignment of a graph is a function with domain such that for each vertex , is a subset of . This set is called the list of admissible colours for . If for each , then is a list assignment. The size of a list assignment is the maximum list size over all vertices . A colouring respects if for all . Given a graph with a list assignment , the List Colouring problem is to decide whether has a colouring that respects . If is fixed, then we have the List Colouring problem. Fixing the size of to be at most gives the List Colouring problem. We say that a list assignment of a graph is regular if, for all , contains exactly colours. This gives us the following problem, which is one focus of this paper. It is defined for each integer (note that is fixed; that is, is not part of the input).
Regular List Colouring
Instance: a graph with an regular list assignment .
Question: does have a colouring that respects ?
A precolouring of a graph is a function for some subset . A colouring of is an extension of a precolouring of if for each . Given a graph with a precolouring , the Precolouring Extension problem is to decide whether has a colouring that extends . If is fixed, we obtain the Precolouring Extension problem.
The relationships amongst the problems introduced are shown in Figure 1.
For an integer , a graph is choosable if, for every regular list assignment of , there exists a colouring that respects . The corresponding decision problem is the Choosability problem. If is fixed, we obtain the Choosability problem.
We emphasize that Regular List Colouring and Choosability are two fundamentally different problems. For the former we must decide whether there exists a colouring that respects a particular regular list assignment. For the latter we must decide whether or not every regular list assignment has a colouring that respects it. As we will see later, this difference also becomes clear from a complexity point of view: for some graph classes Regular List Colouring is computationally easier than Choosability, whereas, perhaps more surprisingly, for other graph classes, the reverse holds.
For two vertexdisjoint graphs and , we let denote the disjoint union , and denote the disjoint union of copies of . If is a graph containing an edge or a vertex then and denote the graphs obtained from by deleting or , respectively. If is a subgraph of then denotes the graph with vertex set and edge set . We let , and denote the cycle, complete graph and path on vertices, respectively. A wheel is a cycle with an extra vertex added that is adjacent to all other vertices. The wheel on vertices is denoted ; note that . A graph on at least three vertices is connected if it is connected and there is no vertex whose removal disconnects it. A block of a graph is a maximal subgraph that is connected and cannot be disconnected by the removal of one vertex (so a block is either connected a or an isolated vertex). A block graph is a connected graph in which every block is a complete graph. A Gallai tree is a connected graph in which every block is a complete graph or a cycle. We say that is a leafblock of a connected graph if contains exactly one cut vertex of and is a component of . For a set of graphs , a graph is free if contains no induced subgraph isomorphic to a graph in , whereas is subgraphfree if it contains no subgraph isomorphic to a graph in . The girth of a graph is the length of its shortest cycle.
1.2 Known Results for Planar Graphs
Garey et al. proved the following result, which is in contrast to the fact that every planar graph is colourable by the Four Colour Theorem [2].
Theorem 1.2 ([14])
Colouring is NPcomplete for planar graphs of maximum degree .
Next we present results found by several authors on the existence of choosable graphs for various graph classes.
Theorem 1.3
The following statements hold for choosability:

Every planar graph is choosable [25].

There exists a planar graph that is not choosable [29].

Every planar trianglefree graph is choosable [20].

Every planar graph with no cycles is choosable [21].

There exists a planar trianglefree graph that is not choosable [30].

There exists a planar graph with no cycles, no cycles and no intersecting triangles that is not choosable [24].

Every planar bipartite graph is choosable [1].
We note that smaller examples of graphs than were used in the original proofs have been found for Theorems 1.3.(ii) [18], 1.3.(v) [23] and 1.3.(vi) [34] and that Theorem 1.3.(vi) strengthens a result of Voigt [31]. We recall that Thomassen [26] first showed that every planar graph of girth at least is choosable, and that a number of results have since been obtained on choosability of planar graphs in which certain cycles are forbidden; see, for example, [7, 11, 32, 33].
We will also use the following result of Chlebík and Chlebíková.
Theorem 1.4 ([8])
List Colouring is NPcomplete for regular planar bipartite graphs that have a list assignment in which each list is one of , , , and all the neighbours of each vertex with three colours in its list have two colours in their lists.
1.3 New Results for Planar Graphs
Theorems 1.1–1.3 have a number of immediate consequences for the complexity of Regular List Colouring when restricted to planar graphs. For instance, Theorem 1.2 implies that 3Regular List Colouring is NPcomplete for planar graphs, whereas Theorem 1.3.(i) shows that 5Regular List Colouring is polynomialtime solvable on this graph class. As such, it is a natural question to determine the complexity for the missing case . In this section we settle this missing case and also present a number of new hardness results for Regular List Colouring restricted to various subclasses of planar graphs. At the end of this section we show how to combine the known results with our new ones to obtain a number of dichotomy results (Corollaries 3–6). We deduce some of our new results from two more general theorems, namely Theorems 1.5 and 1.6, which we state below, but which we prove in Sections 2.1 and 2.2, respectively.
Theorem 1.5
Let be a finite set of connected planar graphs. Then Regular List Colouring is NPcomplete for planar subgraphfree graphs if there exists a planar subgraphfree graph that is not choosable.
Note that the class of subgraphfree graphs is contained in the class of free graphs. Hence, whenever a problem is NPcomplete for subgraphfree graphs, it is also NPcomplete for free graphs.
Combining Theorem 1.5 with Theorem 1.3.(ii) yields the following result which, as we will see later, was the only case for which the complexity of Regular List Colouring for planar graphs was not settled.
Corollary 1
Regular List Colouring is NPcomplete for planar graphs.
Theorem 1.5 has more applications. For instance, consider the non4choosable planar graph from the proof of Theorem 1.7 in [18]. It can be observed that is subgraphfree for all . Wheels are connected and planar. Hence if is any finite set of wheels on at least eight vertices then Regular List Colouring is NPcomplete for planar subgraphfree graphs.
Our basic idea for proving Theorem 1.5 is to pick a minimal counterexample with list assignment (which may exist due to Theorem 1.3.(ii)). We select an “appropriate” edge and consider the graph . We reduce from an appropriate colouring problem restricted to planar graphs and use copies of as a gadget to ensure that we can enforce a regular list assignment. The proof of the next theorem also uses this idea.
Theorem 1.6
Let be a finite set of connected planar graphs. Then Regular List Colouring is NPcomplete for planar subgraphfree graphs if there exists a planar subgraphfree graph that is not choosable.
Theorem 1.6 has a number of applications. For instance, if we let then Theorem 1.6, combined with Theorem 1.3.(v), leads to the following result.
Corollary 2
Regular List Colouring is NPcomplete for planar trianglefree graphs.
Theorem 1.6 can also be used for other classes of graphs. For example, let be a finite set of graphs, each of which includes a connected graph on at least five vertices as a subgraph. Let be the set of these connected graphs. The graph is a planar subgraphfree graph that is not choosable (since it is not colourable). Therefore, Theorem 1.6 implies that Regular List Colouring is NPcomplete for planar subgraphfree graphs. We can obtain more hardness results by taking some other planar graph that is not choosable, such as a wheel on an even number of vertices. Also, if we let we can use Theorem 1.6 by combining it with Theorem 1.3.(vi) to find that Regular List Colouring is NPcomplete for planar graphs with no cycles and no cycles. We strengthen this result as follows (see Section 2.3 for the proof).
Theorem 1.7
Regular List Colouring is NPcomplete for planar graphs with no cycles, no cycles and no intersecting triangles.
Corollaries 1 and 2 and Theorem 1.7 can be seen as strengthenings of Theorems 1.3.(ii), 1.3.(v) and 1.3.(vi), respectively. Moreover, they complement Theorem 1.2, which implies that List Colouring is NPcomplete for planar graphs, and a result of Kratochvíl [19] that, for planar bipartite graphs, Precolouring Extension is NPcomplete. Corollaries 1 and 2 also complement results of Gutner [18] who showed that Choosability and Choosability are complete for planar trianglefree graphs and planar graphs, respectively. However, we emphasize that, for special graph classes, it is not necessarily the case that Choosability is computationally harder than Regular List Colouring. For instance, contrast the fact that Choosability is polynomialtime solvable on free graphs [15] with our next result, which we prove in Section 2.4.
Theorem 1.8
Regular List Colouring is NPcomplete for free graphs.
Our new results, combined with known results, close a number of complexity gaps for the Regular List Colouring problem. Combining Corollary 1 with Theorems 1.1, 1.2 and 1.3.(i) gives us Corollary 3. Combining Theorem 1.7 with Theorems 1.1 and 1.3.(iv) gives us Corollary 4. Combining Corollary 2 with Theorems 1.1 and 1.3.(iii) gives us Corollary 5, whereas Theorems 1.1 and 1.3.(vii) imply Corollary 6.
Corollary 3
Let be a positive integer. Then Regular List Colouring, restricted to planar graphs, is NPcomplete if and polynomialtime solvable otherwise.
Corollary 4
Let be a positive integer. Then Regular List Colouring, restricted to planar graphs with no cycles and no cycles and no intersecting triangles, is NPcomplete if and polynomialtime solvable otherwise (even if we allow intersecting triangles and cycles).
Corollary 5
Let be a positive integer. Then Regular List Colouring, restricted to planar trianglefree graphs, is NPcomplete if and polynomialtime solvable otherwise.
Corollary 6
Let be a positive integer. Then Regular List Colouring, restricted to planar bipartite graphs, is polynomialtime solvable.
1.4 Known Results for Bounded Degree Graphs
First we present a result of Kratochvíl and Tuza [20].
Theorem 1.9 ([20])
List Colouring is polynomialtime solvable on graphs of maximum degree at most .
Brooks’ Theorem [6] states that every graph with maximum degree has a colouring unless
is a complete graph or a cycle with an odd number of vertices. The next result of Vizing
[28] generalizes Brooks’ Theorem to list colourings.Theorem 1.10 ([28])
Let be a positive integer. Let be a connected graph of maximum degree at most and let be a regular list assignment for . If is not a cycle or a complete graph then has a colouring that respects .
And we need another result of Chlebík and Chlebíková [8].
Theorem 1.11 ([8])
Precolouring Extension is polynomialtime solvable on graphs of maximum degree .
1.5 New Results for Bounded Degree Graphs
In Section 2.5, we prove the following result by making a connection to Gallai trees.
Theorem 1.12
Let be a positive integer. Then Precolouring Extension is polynomialtime solvable for graphs of maximum degree at most .
We have the following two classifications. The first one is an observation obtained by combining only previously known results, whereas the second one also makes use of our new result.
Corollary 7
Let be a positive integer. The following two statements hold for graphs of maximum degree at most .

List Colouring is NPcomplete if and polynomialtime solvable if .

Precolouring Extension and Colouring are NPcomplete if and polynomialtime solvable if .
Proof
Corollary 8
Let and be two positive integers. The following two statements hold for graphs of maximum degree at most .

List Colouring and List Colouring are NPcomplete if and and polynomialtime solvable otherwise.

Regular List Colouring and Precolouring Extension are NPcomplete if and and polynomialtime solvable otherwise.
Proof
We first consider (i). If and , we use Theorem 1.4. If or , we use Theorems 1.1 or 1.9, respectively.
We now consider (ii). We start with the hardness cases and so let and .
First consider Precolouring Extension. Theorem 1.2 implies that Colouring is NPcomplete for graphs of maximum degree at most for all . The case follows immediately from this result. Suppose and . Consider a graph of maximum degree . For each vertex , we add new vertices and edges . Let be the resulting graph. Note that has maximum degree at most . We define a precolouring on the newly added vertices by assigning colour to each . Then has a colouring extending if and only if has a colouring.
Now consider Regular List Colouring. The case follows immediately from Theorem 1.2. Suppose and . Consider a graph of maximum degree . We define the list for each vertex . For each vertex , we add new vertices and edges . We define the list for each . For each vertex , we also add new vertices and edges such that form a clique (on vertices). We define the list for each . Let be the resulting graph. Note that has maximum degree at most and that the resulting list assignment is a regular list assignment of . Then has a colouring respecting if and only if has a colouring.
Note that Corollary 8 does not contain a dichotomy for Colouring restricted to graphs of maximum degree at most . A full classification of this problem is open, but a number of results are known. Molloy and Reed [22]classified the complexity for all pairs for sufficiently large . EmdenWeinert et al. [12] proved that Colouring is NPcomplete for graphs of maximum degree at most .
2 Proofs
2.1 The Proof of Theorem 1.5
We need an additional result.
Theorem 2.1
For every integer , List Colouring is NPcomplete for planar graphs of girth at least that have a list assignment in which each list is one of , , , .
Proof
By Theorem 1.4, List Colouring is NPcomplete for regular planar bipartite graphs that have a list assignment in which each list is one of , , , and all the neighbours of each vertex with three colours in its list have two colours in their lists. We modify the hardness construction as follows. Note that for each edge at least one of the incident vertices has a list of size . We replace each edge by a path on an odd number of edges in such a way that the girth of the graph obtained is at least . The new vertices on the path are all given the same list of size , identical to the list on one or other of the endvertices. It is readily seen that these modifications do not affect whether or not the graph can be coloured.∎
We are now ready to prove Theorem 1.5, which we restate below.
Theorem 1.5 (restated). Let be a finite set of connected planar graphs. Then Regular List Colouring is NPcomplete for planar subgraphfree graphs if there exists a planar subgraphfree graph that is not choosable.
Proof
The problem is readily seen to be in NP. Let be a planar subgraphfree graph with a regular list assignment such that has no colouring respecting . We may assume that is minimal (with respect to the subgraph relation). In particular, this means that is connected. Let be the length of a longest cycle in any graph of . We reduce from the problem of List Colouring restricted to planar graphs of girth at least in which each vertex has list , , or . This problem is NPcomplete by Theorem 2.1. Let a graph and list assignment be an instance of this problem. We will construct a planar subgraphfree graph with a regular list assignment such that has a colouring that respects if and only if has a colouring that respects .
If every pair of adjacent vertices in has the same list, then the problem of finding a colouring that respects is just the problem of finding a colouring which, by the Four Colour Theorem [2], we know is possible. Thus we may assume that, on the contrary, there is an edge such that . Let . Then, by minimality, has at least one colouring respecting , and moreover, for any colouring of that respects , and are coloured alike (otherwise we would have a colouring of that respects ). Let be the set of possible colours that can be used on and in colourings of that respect and let . As , we have . Up to renaming the colours in , we can build copies of with regular list assignments such that

the set is any given list of colours of size , and

the vertex corresponding to has any given list of colours containing .
We will implicitly make use of this several times in the remainder of the proof.
We say that a vertex in is a bivertex or trivertex if is or , respectively. We construct a planar subgraphfree graph and regular list assignment as follows.
First suppose that . For each bivertex in , we do as follows. We add two copies of to , which we label and . The vertex in corresponding to is labelled for and we set . We add the edges and . We give list assignments to the vertices of and such that for and for . We let . For each trivertex in , we do as follows. We add one copy of to , which we label . The vertex in corresponding to is labelled and we set . We add the edge . We give list assignments to vertices of such that for . We let . This completes the construction of and when .
Now suppose that . Let if is even and if is odd (so is even in both cases). For each bivertex in , we do as follows. We add a copy of to , which we label , and identify the vertex in corresponding to with . We give list assignments to vertices of such that and . For each trivertex in , we do as follows. We add copies of to which we label , . The vertex in corresponding to is labelled . Let . Add edges such that the union of and induces a cycle on vertices. For all , we give list assignments to vertices of such that . We let . This completes the construction of and when .
Now suppose that . For each bivertex in , we do as follows. We add two copies of to which we label and , such that for , the vertex in corresponding to is identified with . We give list assignments to vertices of and such that for , for and . For each trivertex in , we do as follows. We add a copy of to which we label , such that the vertex in corresponding to is identified with . We give list assignments to the vertices of such that and . This completes the construction of and when .
Note that is planar. Suppose that there is a subgraph in that is isomorphic to a graph of . Since is subgraphfree, and since is obtained from by removing one edge, is also subgraphfree. Therefore for all , is not fully contained in any . Since is connected and since for all only one vertex of any has a neighbour outside of , we find that has at most one vertex in each . In particular, cannot contain any vertex of any in which the vertex corresponding to has been attached to (as opposed to being identified with ); this includes the case when the union of and induces a cycle on vertices. Hence we have found that is a subgraph of , which contradicts the fact that has girth at least . Therefore is subgraphfree.
Note that in any colouring of that respects , each copy of must be coloured such that the vertices corresponding to and have the same colour, which must be one of the colours from the corresponding set . If and is a trivertex, this means that the unique neighbour of in must be coloured with colour , so cannot be coloured with colour . Similarly, if and is a bivertex or and is a trivertex then the two neighbours of in must be coloured with colours and , so cannot be coloured with colours or . If and is a bivertex or and is a trivertex then belongs to a copy of with , so cannot have colour or . If and is a bivertex then belongs to two copies of , one with and one with . Therefore, must be coloured with a colour from the intersection of these two sets, that is it must be coloured with a colour from . Therefore none of the vertices of can be coloured or . Thus the problem of finding a colouring of that respects is equivalent to the problem of finding a colouring of that respects . This completes the proof.∎
2.2 The Proof of Theorem 1.6
Theorem 1.6 (restated). Let be a finite set of connected planar graphs. Then Regular List Colouring is NPcomplete for planar subgraphfree graphs if there exists a planar subgraphfree graph that is not choosable.
Proof
The problem is readily seen to be in NP. Every graph in is connected, and therefore contains a cycle. Let be the length of a longest cycle in any graph of . By assumption, there exists a planar graph and regular list assignment such that is subgraphfree and has no colouring respecting . We may assume that is minimal by removing edges and vertices until any further removal would give a graph with a colouring respecting . In particular, we note that is connected.
We distinguish two cases.
Case 1. is the same for every vertex in .
Then we may assume without loss of generality that for all .
We reduce from Colouring which is NPcomplete even for planar graphs by Theorem 1.2.
Let be a planar graph.
We will construct a planar subgraphfree graph as follows.
Let be an edge of . Let . Then, by minimality, has at least one colouring respecting , which must be a colouring as every list consists of the colours . For every colouring of , it holds that (otherwise would be a colouring of that respects ). Moreover, since we can permute the colours, there is such a colouring that colours (and thus ) with colour for each . Note that in the vertices and must be at distance at least from each other.
Let . Assume that the vertices of are ordered. For each edge with , we do the following:

delete ;

add copies of labelled and, for , let and be the vertices in corresponding to and ;

identify with and, for , identify with ;

add an edge from to .
Let be the obtained graph and note that is planar. Every cycle in that is not contained in a copy of has length at least , since it corresponds to a cycle in and in which every edge has been replaced by successive copies of plus an edge.
Suppose that there is a subgraph in that is isomorphic to a graph of . Since is subgraphfree, and since is obtained from by removing one edge, must also be subgraphfree. Therefore is not fully contained in a copy of . Since is connected, this implies that there is a cycle in that is not fully contained in a copy of . By definition, this cycle has length at most , a contradiction.
Suppose the graph has a colouring . For each copy of , the vertices corresponding to and must be coloured the same. For all edges with in , there is a vertex in coloured the same as in that is adjacent to in , so . Therefore restricted to is a colouring of .
On the other hand, suppose the graph has a colouring . We can extend this colouring to by doing the following: for all edges with in , colour every in such a way that the vertex corresponding to and the vertex corresponding to have colour . This leads to a colouring of .
Case 2. contains two vertices and with .
As is connected, we assume without loss of generality that and are adjacent; let .
We reduce from the problem of List Colouring restricted to planar graphs of girth at least in which each vertex has list , , or . This problem is NPcomplete by Theorem 2.1. Let a graph and list assignment be an instance of this problem. We will construct a planar subgraphfree graph with a regular list assignment such that has a colouring that respects if and only if has a colouring that respects .
We define . Then, by minimality, has at least one colouring respecting , and moreover, for any colouring of that respects , and are coloured alike (otherwise we would have a colouring of that respects ). Let be the set of possible colours that can be used on and in colourings of that respect and let . As , we have . Let us assume, without loss of generality, that and that .
We say that a vertex in is a bivertex or trivertex if is or , respectively. We construct a planar free graph .
First suppose that . For each bivertex in , we do as follows. We add a copy of to which we label . The vertex in corresponding to is labelled and we set . We add the edge . This completes the construction of when .
Now suppose that . Let if is even and if is odd (so is even in both cases). For each bivertex in , we add copies of to which we label , . The vertex in corresponding to is labelled . Let . Add edges such that, for each bivertex in , the union of and induces a cycle on vertices. This completes the construction of when .
Note that is planar, since it is made of planar graphs ( and copies of ) connected in a way that does not obstruct planarity. Suppose that there is a subgraph in that is isomorphic to a graph of . Since is subgraphfree, and since is obtained from by removing one edge, is also subgraphfree. Therefore for all , is not fully contained in . Since is connected and since for all , only one vertex of has a neighbour outside of , we find that has at most one vertex in each . This means that is a subgraph of , which contradicts the fact that has girth at least . Therefore is subgraphfree.
Now we define a list assignment . We give the vertices of each copy of the same lists as their corresponding vertices in , and for each bivertex in , we define , and for each trivertex in , we define . This gives us the regular list assignment of .
The graph has a colouring that respects (the restriction of) and we notice that in such a colouring each copy of must be coloured in such a way that, for each bivertex in , the vertices of that are adjacent to are coloured with the colours of . So one of the neighbours of in must be coloured . Thus the problem of finding a colouring of that respects is equivalent to the problem of finding a colouring of that respects . The proof is complete.∎
2.3 The Proof of Theorem 1.7
Theorem 1.7 is not quite implied by Theorem 1.6. However, we can adapt the proof of Theorem 1.6 to prove Theorem 1.7, which we restate below.
Theorem 1.7 (restated). Regular List Colouring is NPcomplete for planar graphs with no cycles, no cycles and no intersecting triangles.
Proof
Let be the graph on five vertices with two triangles sharing exactly one vertex (this graph is known as the butterfly). Consider the previous proof with . Note that the only problem is that is not connected.
Consider the example in [24] of a graph with no cycle, no cycle and no intersecting triangles that is not choosable. Let be this example, with a regular list assignment such that there is no colouring of respecting . We remove edges and vertices from until any further removal would give a graph with a colouring respecting . This leads to a connected graph . There are no four vertices in with the same list inducing a connected subgraph, so there are no such four vertices in . Therefore in there is an edge such that (otherwise would have at most three vertices, and thus would be choosable).
Therefore we can skip Case and directly adapt the proof in Case . Note that the only thing to prove is that in this case does not contain a subgraph isomorphic to . Suppose is such a subgraph. Since is subgraphfree, cannot be fully contained in for any . Since no vertex is in two different and no vertex of has two adjacent neighbours outside , this implies that there is a triangle in , which is impossible since has girth at least .∎
2.4 The Proof of Theorem 1.8
The proof is obtained by a modification of the NPhardness construction of List Colouring for free graphs from [17]. Recall that we included this result in our paper to illustrate that Choosability and Regular List Colouring can have different complexities when restricted to special graph classes. Indeed, since Choosability is polynomialtime solvable on free graphs [15], Theorem 1.8 shows that Choosability may even be easier than Regular List Colouring.
Theorem 1.8 (restated). Regular List Colouring is NPcomplete for free graphs.
Proof
The problem is readily seen to belong to NP. Golovach et al. [15] showed that List Colouring is NPcomplete for free graphs in which every vertex has a list of size or . Let be such an instance. We add three new vertices to . We make adjacent to each other and to each original vertex of . This results in a free graph . We take three new colours and set . This forces colour to be used to colour one of or . Then all that remains is to add colour to the list of every vertex of that has a list of size .∎
2.5 The Proof of Theorem 1.12
We need some additional results. We begin with a theorem of Bonomo et al.
Theorem 2.2 ([3])
List Colouring is polynomialtime solvable on block graphs.
By generalizing their proof we extend this result to classes of graphs where List Colouring is polynomialtime solvable on the blocks of graphs in the class.
Theorem 2.3
Let be a class of graphs and let be the class of graphs that appear as blocks of graphs in . If List Colouring is polynomialtime solvable on then it is polynomialtime solvable on .
Proof
Let be a graph in that, together with a list assignment , forms an instance of List Colouring. We may assume is connected. If then consider a cutvertex in a leafblock (such and exist). Let be the restriction of to . For each colour , we do as follows. We remove from the list of every neighbour of in and check whether admits a colouring that respects . Note that we can do this in polynomial time, as is in . If so then we put in a set for vertex ; otherwise we do not do this. If, after considering each colour in , we find that then we return no. Otherwise we define a new list assignment for the subgraph of induced by by setting and if . Note that has a colouring that respects if and only if has a colouring that respects . We continue with the pair . We do this exhaustively until we obtain in polynomial time a graph in . Since List Colouring is polynomialtime solvable in , this completes the proof.∎
Corollary 9
List Colouring is polynomialtime solvable on Gallai trees.
We also state the following theorem, proved independently by Borodin and Erdős et al.
Theorem 2.4 ([4, 5, 13])
Let be a connected graph with a list assignment such that for all . If is not a Gallai tree then has a colouring respecting .
We now restate and prove Theorem 1.12.
Theorem 1.12 (restated). Let be a positive integer. Then Precolouring Extension is polynomialtime solvable for graphs of maximum degree at most .
Proof
Let , together with a precolouring defined on a subset , be an instance of Precolouring Extension. We let be the subgraph of induced by . For each , we set . Observe that has a colouring extending if and only if has a colouring that respects . Hence we may consider instead.
Note that, for every vertex , the colours that are removed from to obtain are exactly the colours of those neighbours of in that are not in . This implies, together with the assumption that has maximum degree at most , that every has at most neighbours in .
We now apply the following procedure on exhaustively. If a vertex has fewer than neighbours then remove from . In the end we obtain a graph with the property that for all . Moreover, has a colouring that respects if and only if has a colouring that respects the restriction of to . Hence we may consider instead. We consider every connected component of in turn. If is a Gallai tree then we apply Corollary 9. Otherwise we apply Theorem 2.4.∎
3 Conclusions
As well as filling the complexity gaps of a number of colouring problems for graphs with bounded maximum degree, we have given several dichotomies for the Regular List Colouring problem restricted to subclasses of planar graphs. In particular we showed NPhardness of the cases and restricted to planar subgraphfree graphs for several sets of connected planar graphs. Our method implies that for such sets it suffices to find a counterexample to choosability or to choosability, respectively. It is a natural question whether we can determine the complexity of Regular List Colouring and Regular List Colouring for any class of planar subgraphfree graphs. However, we point out that even when restricting to be a finite set of connected planar graphs, this would be very hard (and beyond the scope of this paper) as it would require solving several longstanding conjectures in the literature. For example, when , Montassier [23] conjectured that every planar subgraphfree graph is choosable.
A drawback of our method is that we need the set of graphs to be connected. If we forbid a set of graphs that are not connected, the distinction between polynomialtime solvable and NPcomplete cases is not clear, and both cases may occur even if we forbid only one graph. We illustrate this below with an example.
Example. Let contain only the star for some . Note that subgraphfree graphs are exactly those graphs that have maximum degree at most . Hence, if , then Regular List Colouring is polynomialtime solvable due to Theorem 1.9. However, there exist larger values of for which the problem is NPcomplete. In order to see this we adapt the proof of Theorem 1.6. The hardness reductions in this proof multiply the maximum degree of our instances by some constant that is at most the maximum degree of the noinstance . By Theorems 1.2 and 1.4, the problems we reduce from are NPcomplete even for graphs with maximum degree at most . Hence, we have proven the following: if is a finite set of connected planar graphs and is a nonchoosable planar subgraphfree graph with maximum degree , then Regular List Colouring is NPcomplete on planar subgraphfree graphs with maximum degree at most . We can take to deduce that Regular List Colouring is NPcomplete on planar subgraphfree graphs.
Acknowledgements. We thank Steven Kelk for helpful comments on an earlier version of this paper.
References
 [1] N. Alon and M. Tarsi, Colorings and orientations of graphs, Combinatorica 12 (1992) 125–134.
 [2] K. Appel and W. Haken, Every planar map is four colorable, Contemporary Mathematics 89, AMS Bookstore, 1989.
 [3] F. Bonomo, G. Durán and J. Marenco, Exploring the complexity boundary between coloring and listcoloring, Ann. Oper. Res. 169 (2009) 3–16.
 [4] O.V. Borodin, Criterion of chromaticity of a degree prescription, in: Abstract of IV AllUnion Conf. on Theoretical Cybernetics (Novosibirsk) (1977) 127128 (in Russian).
 [5] O.V. Borodin, Problems of coloring and of covering the vertex set of a graph by induced subgraphs. Ph.D. Thesis, Novosibirsk, 1979 (in Russian).
 [6] R. L. Brooks, On colouring the nodes of a network, Mathematical Proceedings of the Cambridge Philosophical Society 37 (1941) 194–197.
 [7] M. Chen, M. Montassier and A. Raspaud, Some structural properties of planar graphs and their applications to 3choosability, Discrete Mathematics 312 (2012) 362–373.
 [8] M. Chlebík and J. Chlebíková, Hard coloring problems in low degree planar bipartite graphs, Discrete Applied Mathematics 154 (2006) 1960–1965.
 [9] M. Chudnovsky, Coloring graphs with forbidden induced subgraphs, Proc. ICM 2014 vol IV, 291–302.
 [10] K.K. Dabrowski, F. Dross, M. Johnson and D. Paulusma, Filling the complexity gaps for colouring planar and bounded degree graphs, Proc. IWOCA 2015, LNCS, to appear.
 [11] Z. Dvořák, B. Lidický and R. Škrekovski, Planar graphs without 3, 7, and 8cycles are 3choosable, Discrete Mathematics 309 (2009) 5899–5904.

[12]
T. EmdenWeinert, S. Hougardy and B. Kreuter, Uniquely colourable graphs and the hardness of colouring graphs of large girth, Combinatorics, Probability & Computing 7 (1998) 375–386.
 [13] P. Erdős, A. L. Rubin, and H. Taylor, Choosability in graphs, Proceedings of the West Coast Conference on Combinatorics, Graph Theory and Computing (Humboldt State Univ., Arcata, Calif., 1979), Congress. Numer., XXVI, Winnipeg, Man., 1980, Utilitas Math., pp. 125–157.
 [14] M.R. Garey, D.S. Johnson, and L.J. Stockmeyer, Some simplified NPcomplete graph problems, Proc. STOC 1974, 47–63.
 [15] P.A. Golovach, P. Heggernes, P. van ’t Hof and D. Paulusma, Choosability on free graphs, Information Processing Letters 113 (2013) 107–110.
 [16] P.A. Golovach, M. Johnson, D. Paulusma, and J. Song, A survey on the computational complexity of colouring graphs with forbidden subgraphs, Manuscript, arXiv:1407.1482v4.
 [17] P.A. Golovach, D. Paulusma and J. Song, Closing complexity gaps for coloring problems on free graphs, Information and Computation 237 (2014) 204–214.
 [18] S. Gutner, The complexity of planar graph choosability, Discrete Mathematics 159 (1996) 119–130.
 [19] J. Kratochvíl, Precoloring extension with fixed color bound. Acta Mathematica Universitatis Comenianae 62 (1993) 139–153.
 [20] J. Kratochvíl and Z. Tuza, Algorithmic complexity of list colourings, Discrete Applied Mathematics 50 (1994) 297–302.
 [21] P.C.B. Lam, B. Xu and J. Liu, The 4choosability of plane graphs without 4cycles, Journal of Combinatorial Theory, Series B 76 (1999) 117–126.
 [22] M. Molloy and B. Reed, Colouring graphs when the number of colours is almost the maximum degree, Journal of Combinatorial Theory, Series B 109 (2014) 134–195.
 [23] M. Montassier, A note on the not 3choosability of some families of planar graphs, Information Processing Letters 99 (2006) 68–71.
 [24] M. Montassier, A. Raspaud and W. Wang, Bordeaux 3color conjecture and 3choosability, Discrete Mathematics 306 (2006) 573–579.
 [25] C. Thomassen, Every planar graph is choosable, J. Combin. Theory Ser. B 62 (1994) 180–181.
 [26] C. Thomassen, 3Listcoloring planar graphs of girth 5, J. Comb. Theory, Ser. B 64 (1995) 101–107.
 [27] V.G. Vizing, Coloring the vertices of a graph in prescribed colors, in Diskret. Analiz., no. 29, Metody Diskret. Anal. v. Teorii Kodov i Shem 101 (1976) 3–10.
 [28] V.G. Vizing, Vertex colorings with given colors, Diskret. Analiz. 29 (1976) 3–10.
 [29] M. Voigt, List colourings of planar graphs, Discrete Mathematics 120 (1993) 215–219.
 [30] M. Voigt, A not 3choosable planar graph without 3cycles, Discrete Mathematics 146 (1995) 325–328.
 [31] M. Voigt, A non3choosable planar graph without cycles of length 4 and 5, Discrete Mathematics 307 (2007) 1013–1015.
 [32] Y. Wang, H. Lu and M. Chen, Planar graphs without cycles of length 4, 5, 8, or 9 are 3choosable. Discrete Mathematics 310 (2010) 147–158.
 [33] Y. Wang, H. Lu and M. Chen, Planar graphs without cycles of length 4, 7, 8, or 9 are 3choosable. Discrete Applied Mathematics 159 (2011) 232–239.
 [34] DQ. Wang, YP. Wen and KL. Wang, A smaller planar graph without 4, 5cycles and intersecting triangles that is not 3choosable Information Processing Letters 108 (2008) 87–89.